Paper II — Q5
(a) A one-hour unit hydrograph of a catchment is shown in the figure. A storm of two hours duration with intensity of 70 mm/h in…
A one-hour unit hydrograph of a catchment is shown in the figure. A storm of two hours duration with intensity of 70 mm/h in the first hour and 40 mm/h in the second hour occurs over the catchment. Calculate the peak direct surface runoff value in m³/s. 10 marks
A temporary coffer dam is to be built to protect the 5 year construction activity for a major river valley project. If the coffer dam is designed to withstand the 25-year flood, what is the risk that the structure will be overtopped in the third year? 10 marks
A weir across an alluvial river has a horizontal floor length of 50 m and retains 5 m of water under full flow condition. The downstream sheet pile is driven to a depth of 5 m below the impervious floor of negligible thickness. Determine the exit gradient. 10 marks
What is an indicator organism ? Discuss the required characteristics for an ideal indicator organism. Name any two indicator organisms. 10 marks
Circular sewer of minimum diameter of 150 mm is recommended for the house connections. Calculate the discharge carried by this sewer, when flowing full, laid at a slope of 0·004. Take Manning's n as 0·013.
Why this sewer size has been recommended as the minimum size, when the waste water discharge from individual household will be substantially less ? 10 marks
हिंदी में प्रश्न पढ़ें
एक जलग्रहण क्षेत्र का एक-घंटा एकांकी जलालेख चित्र में दर्शाया गया है । जलग्रहण क्षेत्र पर दो घंटे की अवधि की एक वृष्टि, 70 mm प्रति घंटा की तीव्रता के साथ पहले घंटे में एवं 40 mm प्रति घंटा की तीव्रता के साथ दूसरे घंटे में होती है । चरम प्रत्यक्ष धरातलीय अपवाह के मान की गणना m³/s में कीजिए । (10 अंक)
एक मुख्य नदी घाटी परियोजना की 5 वर्ष की निर्माण गतिविधि की सुरक्षा के लिए एक अस्थायी कॉफर बांध बनाया जाना है । यदि कॉफर बांध का अभिकल्पन 25-साल की बाढ़ को सहने के लिए किया गया है, तो इस संरचना के तीसरे साल में उत्लावन होने का जोखिम कितना है ? (10 अंक)
एक जलोढ़ नदी के आरपार स्थित एक विवर का क्षैतिज फर्श 50 m लंबा है एवं पूर्ण प्रवाह की स्थिति में 5 m पानी रोकता है । अनुप्रवाह (डाउनस्ट्रीम) शीट पाइल, अगम्य मोटाई के अपारगम्य फर्श के नीचे, 5 m गहराई तक डाली गयी है । निर्गम प्रवणता निर्धारित कीजिए । (10 अंक)
एक सूचक जीव क्या है ? एक आदर्श सूचक जीव के लिए आवश्यक अभिलक्षणों की विवेचना कीजिए । किन्हीं दो सूचक जीवों के नाम लिखिए । (10 अंक)
घरेलू संयोजन के लिए न्यूनतम 150 mm व्यास का वृत्ताकार मलक नल (सीवर) अनुशंसित है । इस मलक नल द्वारा प्रवाहित निस्सरण की गणना कीजिए, जब यह पूरा भरा बह रहा है एवं 0·004 के ढाल पर बिछाया गया है । मैनिंग के n को 0·013 लीजिए ।
इस मलक नल के आमाप की अनुशंसा न्यूनतम आमाप के रूप में क्यों की गयी है, जबकि पृथक घर से निस्सरित अपशिष्ट जल प्रवाह अत्यंत ही कम होगा ? (10 अंक)
The figure this question refers to, in words
The question paper is a scan and the diagram did not survive as text. This is the figure as read from the original page — every component, value and label — so the question can be worked from the text below.
(a) A graph of a 1-hour unit hydrograph plotted with Discharge (m3/s) on the vertical y-axis and Time (h) on the horizontal x-axis. The y-axis is marked from 0 to 3 in intervals of 1. The x-axis is marked from 0 to 4 in intervals of 1. The hydrograph is triangular, starting at the origin (0, 0), rising linearly to a peak discharge of 3 m3/s at time t = 1 h, and then falling linearly to 0 m3/s at time t = 4 h. Dashed guide lines show the points along the recession limb: at t = 2 h, discharge = 2 m3/s; at t = 3 h, discharge = 1 m3/s.
Model answer
Drafted with AI by UPSC Answer Check against this question's marking rubric, to the expected length. UPSC does not publish answers for Mains — this is one way to score well, not an official key. How we build rubrics and answers.
(a) Peak direct runoff from the two-hour storm
No losses are given, so the intensities are taken as rainfall excess: P1 = 70 mm = 7 cm and P2 = 40 mm = 4 cm. From the figure, the 1-h unit hydrograph ordinates (m³/s per cm) at t = 0, 1, 2, 3, 4 h are 0, 3, 2, 1, 0.
Direct runoff Q(t) = 7·U(t) + 4·U(t−1):
- t = 1 h: 7×3 = 21 m³/s
- t = 2 h: 7×2 + 4×3 = 26 m³/s
- t = 3 h: 7×1 + 4×2 = 15 m³/s
- t = 4 h: 7×0 + 4×1 = 4 m³/s
Between 1 and 2 h, Q = 16 + 5t (rising); between 2 and 3 h, Q = 48 − 11t (falling). So the peak occurs at t = 2 h.
Peak direct surface runoff = 26 m³/s, at 2 h after the start of the storm.
(b) Risk of overtopping of the coffer dam in the third year
The dam is designed for the 25-year flood, so the annual exceedance probability is p = 1/T = 1/25 = 0.04. Annual floods are assumed independent, so the probability that the dam is overtopped in any one particular year, including the third, is the same.
Risk in the third year = 0.04 = 4%.
For comparison, the cumulative risk of at least one overtopping, R = 1 − (1 − p)ⁿ, is 1 − 0.96³ = 11.5% by the end of the third year and 1 − 0.96⁵ = 18.5% over the full 5-year construction period.
(c) Exit gradient by Khosla's theory
For a downstream pile of depth d at the end of a floor of length b under head H:
G_E = (H/d) × 1/(π√λ), with λ = [1 + √(1 + α²)]/2 and α = b/d.
Here H = 5 m, b = 50 m, d = 5 m, so α = 10.
λ = (1 + √101)/2 = (1 + 10.05)/2 = 5.525, so √λ = 2.351.
G_E = (5/5) × 1/(π × 2.351) = 1/7.384 = 0.135.
Exit gradient ≈ 0.135 (about 1 in 7.4). This is below the usual safe limit of about 0.17 to 0.20 for alluvial soils, so the floor is safe against piping.
(d) Indicator organism and its ideal characteristics
An indicator organism is a micro-organism, usually harmless and easy to count, whose presence in water signals faecal contamination and hence the likely presence of pathogens. It is used because testing for each pathogen is slow and costly.
An ideal indicator should:
- be present in faeces in large numbers, and absent in unpolluted water;
- not multiply outside the intestine, so the count reflects the degree of pollution;
- survive in water at least as long as, and resist disinfection at least as well as, the pathogens;
- be detectable by a simple, quick and cheap test;
- be non-pathogenic, specific to faecal pollution and give consistent results in different waters.
Two indicator organisms: Escherichia coli (the coliform group, counted by MPN) and faecal streptococci such as Enterococcus faecalis.
(e) Discharge of a 150 mm sewer and the reason for the minimum size
For a pipe flowing full: A = πD²/4 = π(0.15)²/4 = 0.01767 m², and R = D/4 = 0.0375 m.
Manning: Q = (1/n)·A·R^(2/3)·S^(1/2) = (1/0.013) × 0.01767 × 0.1120 × 0.0632 = 0.0096 m³/s.
Q ≈ 0.0096 m³/s ≈ 9.6 L/s, with full-flow velocity V = Q/A ≈ 0.55 m/s.
The minimum size is fixed by blockage, not by flow. The solids, paper, rags and grease in domestic sewage choke smaller pipes, particularly at bends and junctions and in the long periods of low flow. A 150 mm pipe passes such solids, can be rodded and cleaned, and keeps spare capacity for peak flows, infiltration and later additions. A pipe sized only for the very small household discharge would clog and be hard to maintain.
How this answer will be evaluated
Framework: null
What earns full marks: Accurate calculations with clear steps, correct formulas, and well-reasoned explanations for all parts.
Evaluation rubric
Each sub-part is marked on its own, against the marks and word limit printed on the paper.
- (a) Peak direct surface runoff value in m³/s using the unit hydrograph method. 10 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- Extract UH ordinates from the provided graph
- Apply storm intensities (70 mm/h, 40 mm/h) to UH
- Perform lagged superposition of the two UHs
- Sum ordinates to find the peak value
Loses marks
- Incorrect reading of UH ordinates from graph
- Failure to lag the second UH by 1 hour
Earns more
- Tabular presentation of the superposition
- Clear identification of the peak time
Extra mark
- Sketch of the resulting storm hydrograph
- (b) Risk of overtopping in the third year for a 25-year flood design. 10 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- State the annual exceedance probability (p = 1/25)
- Note that the risk R = 1 − (1 − p)ⁿ applies to cumulative risk, whereas a single named year has risk p
- State the third-year risk as 0.04 (4%), assuming independent annual floods
Loses marks
- Giving the cumulative risk 1 − 0.96³ as the answer to 'in the third year' without comment
- Confusing return period with probability
Earns more
- Clear definition of the risk variable
- Step-by-step substitution into the formula
Extra mark
- Cumulative risk values (11.5% by year 3, 18.5% over 5 years) shown for comparison
- (c) Exit gradient for the weir using Khosla's theory. 10 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- Identify the parameters (b = 50 m, d = 5 m, H = 5 m)
- Calculate the shape factor (λ) using the standard formula
- Determine the exit gradient (G_e) using the G_e formula
Loses marks
- Incorrect formula for the shape factor
- Confusing head (H) with depth (d)
Earns more
- Recognising the downstream pile at the end of the floor (standard Khosla exit-gradient case)
- Accurate calculation of the shape factor
Extra mark
- Comparison of G_E = 0.135 with the safe exit gradient (about 0.17 to 0.20)
- (d) Definition, characteristics, and examples of indicator organisms. 10 marks
discuss— intro → 3-4 dimensions → example → balanced close
Must cover
- Define 'indicator organism' in the context of water quality
- List at least 3-4 required characteristics of an ideal indicator
- Name two specific indicator organisms (e.g., E. coli)
Loses marks
- Vague definition without context
- Listing characteristics without explanation
Earns more
- Explanation of why the characteristics are necessary
- Distinction between fecal and non-fecal indicators
Extra mark
- Mention of specific testing methods (e.g., MPN)
- (e) Discharge of a 150mm sewer flowing full and justification for minimum size. 10 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- Use Manning's formula for full flow conditions
- Substitute given values (D=0.15m, S=0.004, n=0.013)
- Calculate the discharge in m³/s
- Explain the rationale for the 150mm minimum size
Loses marks
- Incorrect calculation of the hydraulic radius
- Failure to address the 'why' part of the question
Earns more
- Clear statement of the Manning's formula
- Logical explanation of self-cleansing velocity
Extra mark
- Mention of specific IS code for sewer sizing
What "Calculate" is asking you to do
Apply the standard formula or schedule to data the question has already supplied — a table of readings, cost records, a balance sheet — and produce the number. The method is rarely in doubt; the marks sit in the named intermediate quantities, each of which has to appear as a labelled line.
Structure that answers it
Data as given → formula or standard treatment, named → substitution → each intermediate, labelled → result with units
Where marks are lost
Omitting an intermediate the marking scheme pays for separately, or rounding at an intermediate line so the final figure drifts. In commerce and accountancy, any figure in a statement that no numbered working note supports is treated as unearned.
Practice this exact question
Write your answer and it is marked point by point against the model answer above — what you covered, what you missed, what you got wrong.
Evaluate my answer →More from Civil Engineering 2021 Paper II
- Q2 (a) For a small project, the number of masons required is shown. The table also indicates…
- Q3 (a) (i) Explain with neat sketches, how surface and sub-surface water can be removed from…
- Q4 (a) Design the size and spacing of dowel bars at the expansion joints of a cement concret…
- Q5 (a) A one-hour unit hydrograph of a catchment is shown in the figure. A storm of two hour…
- Q6 (a) Two settling tanks – one rectangular, with length to width ratio of 3 : 1 and side wa…
- Q7 (a) An open drain is to be designed to prevent waterlogging for an area of 576 ha. Given…
- Q8 (a) From the basin map of a given watershed with drainage area 3000 km², the length of th…